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\title{人口模型}
\author{五六七}
%\date{2025年9月24日}

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\begin{document}

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  \titlepage
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  \tableofcontents
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\section{Malthus 模型}
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\begin{frame}[allowframebreaks]{Malthus 模型}

{\color{red}问题：建立人口总数关于时间变化的微分方程模型。}

解答：1798年，Malthus 模型，{\color{blue} $\frac{dN}{dt} = rN$ }. 相对增长率 $r$ 保持不变。

{\color{red}问题：求解微分方程 $\frac{dN}{dt} = rN$. }

解答：分离变量法可得。

\begin{eqnarray*}
\frac{dN}{dt} = rN 	
&\Rightarrow & 	\frac{dN}{N} = rdt \\ 
&\Rightarrow &	\int \frac{dN}{N} = \int rdt \\ 
&\Rightarrow & 	\ln N = rt +C \\ 
&\Rightarrow & 	N = e^{rt +C} = Ke^{rt}, 
\end{eqnarray*}

其中 $K=N(0)$ 是初始时刻 $t=0$ 时的人口数量。

\end{frame}

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\section{Logistic 模型}
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\begin{frame}[allowframebreaks]{Logistic 模型}

1838年， (Verhulst) Logistic 模型：{\color{blue} $\frac{dN}{dt} = (a-bN)N$ }. 增长极限为 $a/b$. 


{\color{red}问题：求解微分方程 $\frac{dN}{dt} = (a-bN)N$. }

解答：分离变量法可得
\begin{eqnarray*}
\frac{dN}{dt} = (a-bN)N 
&\Rightarrow& \frac{dN}{(a-bN)N} = dt  \\ 
&\Rightarrow& \int \frac{dN}{(a-bN)N} = \int dt. 
\end{eqnarray*}

计算待定系数 $A,B$ 化简左边的积分表达式
\begin{eqnarray*}
\frac{1}{(a-bN)N} = \frac{A}{a-bN} + \frac{B}{N} 
\quad\Rightarrow\quad B=\frac{1}{a},\, A=\frac{b}{a}. 
\end{eqnarray*}

积分可得 
$$
\int \frac{dN}{(a-bN)N} = -\frac{A}{b}\ln(a-bN) + B\ln N 
= \frac{1}{a}\ln \frac{N}{a-bN}. 
$$

因此微分方程的解为 
\begin{eqnarray*}
\frac{1}{a}\ln \frac{N}{a-bN} = t+C. 
\end{eqnarray*}

两边取指数，可得
\begin{eqnarray*}
\frac{N}{a-bN} = e^Ce^{at}.
\end{eqnarray*}

求出 $N$ 可得
\begin{eqnarray*}
N= \frac{a}{b+Ke^{-at}}.
\end{eqnarray*}
其中 $K=e^{-C}$ 也是任意常数。


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\section{后续问题}
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\begin{frame}{后续问题}

1. 举例说明指数函数与Logistic函数的图像的区别。

2. 设已有历史人口数据，求出参数 $r, a, b, K$ 的估计值。

3. 这个函数表达式是否较好地拟合了历史数据？

4. 这个函数表达式是否较好地预测了将来的数据？

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